Question

在三角形ABC中 ∠A，∠B，∠C所对应的边分别为a,b,c, 求证a²-b²/c²=sin(A-B)/sinC

即求证：在三角形ABC中 ∠A，∠B，∠C所对应的边分别为a,b,c,，a的平方减b的平方比上c的平方等于sin(A-B)比上sinC

Answer 1

(a^2-b^2)/c^2=[(sinA)^2-(sinB)^2]/(sinC)^2
=sin(A+B)sin(A-B)/(sinC)^2
=sin(PI-C)sin(A-B)/(sinC)^2
=sin(A-B)/sinC
其中
Sin[a + b] Sin[a - b]积化和差公式得
=1/2 (-Cos[2 a] + Cos[2 b])余弦二倍角公式得
=Sin[a]^2 - Sin[b]^2

Answer 2

(a^2-b^2)/c^2=(sin^2A-sin^2B)/sin^2C=(sinA+sinB)(sinA-sinB)/sin^2C=2sin[(A+B)/2]cos[(A-B)/2]*2cos[(A+B)/2]sin[(A-B)/2]/sin^2C=sin(A+B)sin(A-B)/sin^2C=sinCsin(A-B)/sin^2C=sin(A-B)/sinC